Evaluating Sine and Cosine at Special Acute Angles - Problem 3

Norm Prokup

Pi over 3 is another special angle. I want to find the cosine and sine of Pi over 3. Now I’ve drawn the angle Pi over 3 on the unit circle point P is the point on the terminal side of Pi over 3 and I need to find the coordinates of Pi over3 in order to find the sine and cosine of the angle.

But I already know the coordinates of point Q which lies on the angle Pi over 6. The sine of Pi over 6 is ½ and the cosine of Pi over 6 is root 3 over 2 and that gives me the x and y coordinates. This is the x coordinate., this is the y coordinate of point Q.

Point Q is a refection of point P along the line y equals, y equals x. you can kind of see that because Pi over 3 is 60 degrees, Pi over 6 is 30 degrees. This angle is 30 degrees so there is symmetry across the diagonal. That means that to get the coordinates of point P all I need to do is interchange the coordinates of point Q. So point P has coordinates ½ root 3 over 2 and that means the cosine of Pi over 3 is ½ and the sine of 1/3 is root 3 over 2.